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If \(n\geqslant0\), as \(a=3\), the last equation would have limited solutions. You should choose the one with the largest \(m\) (that produces the largest \(b=2am\)), so that you can guess the largest divisor of \(b\) that will be equal to \(b\). I get \(m=1\) and \(b=2am=2·3·1=6\), so the largest divisor of \(b\) is \(6\).

If it's possible \(n<0\), there would not be limitation for the value of \(m\) and, in consequence, we could define any \(b\) (and its largest divisor) as large as we wanted just fixing any desired value of \(m\) (that would always have an associated value of \(n=10-3m^2\)).

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